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重拾演算法之劍指Offier——樹的子結構

樹的子結構

題目描述
輸入兩棵二叉樹A,B,判斷B是不是A的子結構。(ps:我們約定空樹不是任意一個樹的子結構)

class TreeNode {
    int val = 0;
    TreeNode left = null;
    TreeNode right = null;

    public TreeNode(int val) {
        this.val = val;

    }
}

public class Solution {

    public boolean HasSubtree(TreeNode root1, TreeNode root2) {
        boolean
flag = false; if (root1 != null && root2 != null){ if (root1.val == root2.val){ flag = doesTree1haveTree2(root1, root2); } if(flag == false){ flag = HasSubtree(root1.left, root2); } if(flag == false
){ flag = HasSubtree(root1.right, root2); } } return flag; } public boolean doesTree1haveTree2(TreeNode root1, TreeNode root2){ if(root2 == null){ return true; } if(root1 == null || root1.val != root2.val){ return
false; } return doesTree1haveTree2(root1.left,root2.left) && doesTree1haveTree2(root1.right,root2.right); } }