剑指Offer之Java算法习题精讲二叉树专项解析剑指Offer之Java算法习题精讲二

题目一
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解法

/** * Definition for a binary tree node. * public class TreeNode { *int val; *TreeNode left; *TreeNode right; *TreeNode() {} *TreeNode(int val) { this.val = val; } *TreeNode(int val, TreeNode left, TreeNode right) { *this.val = val; *this.left = left; *this.right = right; *} * } */class Solution {int ans; int pre; public int getMinimumDifference(TreeNode root) {ans = Integer.MAX_VALUE; pre = -1; method(root); return ans; }public void method(TreeNode root){if(root==null) return; method(root.left); if(pre==-1){pre = root.val; }else{ans = Math.min(ans,root.val-pre); pre = root.val; }method(root.right); }}

题目二

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解法

/** * Definition for a binary tree node. * public class TreeNode { *int val; *TreeNode left; *TreeNode right; *TreeNode() {} *TreeNode(int val) { this.val = val; } *TreeNode(int val, TreeNode left, TreeNode right) { *this.val = val; *this.left = left; *this.right = right; *} * } */class Solution {int ans = 0; public int findTilt(TreeNode root) {method(root); return ans; }public int method(TreeNode root){if(root==null) return 0; int l = method(root.left); int r = method(root.right); ans += Math.abs(l-r); return l+r+root.val; }}

题目三

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【剑指Offer之Java算法习题精讲二叉树专项解析】 解法

/** * Definition for a binary tree node. * public class TreeNode { *int val; *TreeNode left; *TreeNode right; *TreeNode() {} *TreeNode(int val) { this.val = val; } *TreeNode(int val, TreeNode left, TreeNode right) { *this.val = val; *this.left = left; *this.right = right; *} * } */class Solution {public boolean isSubtree(TreeNode root, TreeNode subRoot) {return dfs(root,subRoot); }public boolean dfs(TreeNode root, TreeNode subRoot){if(root==null) return false; return cheack(root,subRoot)||dfs(root.left,subRoot)||dfs(root.right,subRoot); }public boolean cheack(TreeNode root, TreeNode subRoot){if(root==null&&subRoot==null) return true; if(root==null||subRoot==null||root.val!=subRoot.val) return false; return cheack(root.left,subRoot.left)&&cheack(root.right,subRoot.right); }}

题目四

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解法

/** * Definition for a binary tree node. * public class TreeNode { *int val; *TreeNode left; *TreeNode right; *TreeNode() {} *TreeNode(int val) { this.val = val; } *TreeNode(int val, TreeNode left, TreeNode right) { *this.val = val; *this.left = left; *this.right = right; *} * } */class Solution {public boolean isSameTree(TreeNode p, TreeNode q) {if(p==null&&q==null) return true; if(p==null||q==null||q.val!=p.val) return false; return isSameTree(p.left,q.left)&&isSameTree(p.right,q.right); }}

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